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QLDPhysicsSyllabus dot point

Topic 1: Special relativity

Apply the length contraction formula $L = L_0 / \gamma$ and the relativistic momentum formula $p = \gamma m v$ to predict the contraction of moving objects and the momentum of relativistic particles

A focused answer to the QCE Physics Unit 4 dot point on length contraction and relativistic momentum. Defines proper length, applies $L = L_0 / \\gamma$, and contrasts classical $p = mv$ with relativistic $p = \\gamma m v$.

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What this dot point is asking

QCAA wants you to apply the length contraction and relativistic momentum formulas in numerical problems, recognising that both effects scale by the Lorentz factor Ξ³\gamma at high speeds.

Length contraction

A rod at rest in some frame has the proper length L0L_0, measured by an observer in that frame. From a different frame moving relative to the rod at speed vv, the rod is shorter along the direction of motion:

L=L0Ξ³L = \frac{L_0}{\gamma}

where Ξ³=1/1βˆ’v2/c2\gamma = 1/\sqrt{1 - v^2/c^2}.

Direction matters

Length contraction acts only along the direction of motion. A rod's transverse dimensions (perpendicular to its motion) are unaffected. A passing ball would be seen as flattened along its direction of motion but unchanged in the other two dimensions.

Proper length

The proper length is the longer length, measured in the frame where the rod is at rest. Other observers measure a shorter length.

Length contraction is real

Like time dilation, length contraction is not optical illusion. The atmosphere appears thinner to a relativistic muon (in the muon's frame, the atmosphere is contracted; in Earth's frame, time is dilated; both descriptions agree on whether the muon survives to the surface).

Worked example

A spaceship has proper length L0=100L_0 = 100 m. It travels at v=0.80cv = 0.80 c relative to Earth.

Ξ³=1/1βˆ’0.64=1/0.36=1/0.6β‰ˆ1.667\gamma = 1/\sqrt{1 - 0.64} = 1/\sqrt{0.36} = 1/0.6 \approx 1.667.

L=100/1.667β‰ˆ60L = 100 / 1.667 \approx 60 m.

From Earth, the spaceship is 60 m long.

Relativistic momentum

In classical mechanics, momentum is p=mvp = m v. At relativistic speeds, this formula breaks down. The relativistic momentum is:

p=Ξ³mvp = \gamma m v

where Ξ³=1/1βˆ’v2/c2\gamma = 1/\sqrt{1 - v^2/c^2} and mm is the rest mass.

Why the modification

Classical momentum is not conserved under Lorentz transformations between frames. Relativistic momentum is conserved. Several derivations are possible; the simplest is to require that momentum conservation hold in all frames, which forces the Ξ³\gamma factor.

Behaviour at high speeds

At low speeds, Ξ³β‰ˆ1\gamma \approx 1 and pβ‰ˆmvp \approx m v (classical). At high speeds, Ξ³\gamma grows without bound as vβ†’cv \to c. The momentum required to push a particle to a speed approaching cc grows without limit; no particle with rest mass can reach cc.

Worked example

An electron is accelerated to v=0.99cv = 0.99 c. Find its momentum.

Ξ³=7.09\gamma = 7.09 (from above).

p=7.09Γ—(9.11Γ—10βˆ’31)Γ—(0.99Γ—3Γ—108)=7.09Γ—2.71Γ—10βˆ’22β‰ˆ1.92Γ—10βˆ’21p = 7.09 \times (9.11 \times 10^{-31}) \times (0.99 \times 3 \times 10^8) = 7.09 \times 2.71 \times 10^{-22} \approx 1.92 \times 10^{-21} kg m sβˆ’1^{-1}.

The classical value would be 2.71Γ—10βˆ’222.71 \times 10^{-22}, smaller by a factor of Ξ³=7.09\gamma = 7.09.

Consequences of length contraction and relativistic momentum

Particle accelerators. Designing a particle accelerator requires the relativistic momentum formula. Synchrotrons (where charged particles travel in a circle in a magnetic field) must use p=Ξ³mvp = \gamma m v when calculating the magnetic field needed to keep particles in their circular path.

Cosmic rays. Some high-energy cosmic ray particles have Ξ³>1010\gamma > 10^{10} (extraordinarily relativistic). The classical momentum formula is hopelessly wrong; only p=Ξ³mvp = \gamma m v works.

No object with rest mass reaches cc. The relativistic energy E=γmc2E = \gamma m c^2 (covered in the mass-energy dot point) diverges as v→cv \to c. Infinite energy would be required.

Photons. Photons have zero rest mass and travel at cc. They carry momentum p=E/cp = E/c (consistent with p=Ξ³mvp = \gamma m v in an appropriate limit).

Verification

Particle accelerator data. Charged particles in accelerators (LHC, Tevatron, etc.) follow trajectories that match relativistic momentum predictions. Without the Ξ³\gamma correction, the accelerator's magnetic systems would not bend the particles correctly.

Pion decay. High-energy pions in cosmic-ray cascades have lifetimes consistent with time dilation, and trajectories consistent with relativistic momentum.

Common errors

Length contraction direction. Only along the direction of motion. Perpendicular dimensions are unchanged.

Proper length confusion. Proper length is measured in the rest frame of the object (the longer length). Contracted length is measured in any other frame.

Classical momentum applied to relativistic particle. At v/c>0.1v / c > 0.1, the relativistic correction is measurable; above v/c>0.5v / c > 0.5, it is dominant. Use p=Ξ³mvp = \gamma m v whenever in doubt.

Confusing the formula for contraction with dilation. Length: L=L0/Ξ³L = L_0 / \gamma (shorter). Time: t=Ξ³t0t = \gamma t_0 (longer). The factor of Ξ³\gamma goes opposite ways.

In one sentence

In special relativity, an object's length measured by an observer moving relative to it is shorter than its proper length by a factor of Ξ³\gamma along the direction of motion (L=L0/Ξ³L = L_0 / \gamma), and an object's momentum at relativistic speed is greater than the classical value by the same factor (p=Ξ³mvp = \gamma m v); both effects are negligible at everyday speeds but become substantial above about v=0.1cv = 0.1 c and are routinely measured in particle accelerators and cosmic-ray physics.

Past exam questions, worked

Real questions from past QCAA papers on this dot point, with our answer explainer.

2024 QCAA-style4 marksA spaceship of proper length $100$ m travels past Earth at $0.60 c$. (a) Calculate the length of the spaceship measured by an Earth-based observer. (b) Calculate the relativistic momentum of a $1.0$ kg object in the spaceship as measured from Earth.
Show worked answer β†’

(a) Length contraction. Ξ³=1/1βˆ’0.36=1/0.64=1/0.8=1.25\gamma = 1/\sqrt{1 - 0.36} = 1/\sqrt{0.64} = 1/0.8 = 1.25.

L=L0/Ξ³=100/1.25=80L = L_0 / \gamma = 100 / 1.25 = 80 m.

The spaceship is contracted to 80 m as measured from Earth.

(b) Relativistic momentum.

p=Ξ³mv=1.25Γ—1.0Γ—(0.60Γ—3Γ—108)=1.25Γ—1.8Γ—108=2.25Γ—108p = \gamma m v = 1.25 \times 1.0 \times (0.60 \times 3 \times 10^8) = 1.25 \times 1.8 \times 10^8 = 2.25 \times 10^8 kg m sβˆ’1^{-1}.

Markers reward the correct Ξ³\gamma, the contracted length, and the relativistic momentum formula (with the extra factor of Ξ³\gamma compared to classical).

2023 QCAA-style3 marksAn electron in a particle accelerator has speed $0.99 c$. Compare its classical and relativistic momentum (electron mass $9.11 \times 10^{-31}$ kg).
Show worked answer β†’

Classical: pcl=mv=9.11Γ—10βˆ’31Γ—0.99Γ—3Γ—108=2.71Γ—10βˆ’22p_{\text{cl}} = m v = 9.11 \times 10^{-31} \times 0.99 \times 3 \times 10^8 = 2.71 \times 10^{-22} kg m sβˆ’1^{-1}.

Ξ³\gamma at 0.99c0.99 c: 1/1βˆ’0.9801=1/0.0199β‰ˆ7.091/\sqrt{1 - 0.9801} = 1/\sqrt{0.0199} \approx 7.09.

Relativistic: prel=Ξ³mv=7.09Γ—2.71Γ—10βˆ’22β‰ˆ1.92Γ—10βˆ’21p_{\text{rel}} = \gamma m v = 7.09 \times 2.71 \times 10^{-22} \approx 1.92 \times 10^{-21} kg m sβˆ’1^{-1}.

The relativistic momentum is 7.097.09 times the classical value. At 0.99c0.99 c, classical momentum is a substantial underestimate. The discrepancy grows without limit as v→cv \to c.

Markers reward both calculations and the explicit ratio showing the Ξ³\gamma-fold increase.

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